Use of Function Arctan

Time Limit:10000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status Description It's easy to know that arctan(1/2)+arctan(1/3)=arctan(1).The problem is,to some fixed number A,you have to write a program to calculate the minimum sum B+C.A,B and C are all positive integers and satisfy the equation below:

arctan(1/A)=arctan(1/B)+arctan(1/C)

Input The first line conta

ins a integer number T.T lines follow,each contains a single integer A, 1<=A<=60000.

Output T lines,each contains a single integer whic

h denotes to the minimum sum B+C.

Sample Input 1 1

 /*由题意知:B(C-A)=1+AC;B=（1+AC）/(C-A)；B=A+（1+AA）/(C-A);C增B减；（易知B=C时B+C最小（此时B=C=a可能为小数，故只需B--枚举即可！！！））*/

 #include<stdio.h>

 typedef long long LL;

 int main()

 {

     LL B,C,A,i,j,T;

     scanf("%lld",&T);

     for(i=;i<=T;i++)

     {

         scanf("%lld",&A);

         LL tmp=*A+;

         while((tmp--)>A)

         if((A*tmp+)%(tmp-A)==)

         {

             B=(A*tmp+)/(tmp-A);

             printf("%lld\n",tmp+B);

             break;

         }

     }

 }

巴特西

Use of Function Arctan

最新文章

热门文章